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I am cross-posting this question which I posted in math.stackexchange.com since I realized that there are people in Mathoverflow who are not signed-up there.

Edit: For a topological group $G$ and a topological $G$-module $M$, we define the $n$th group of continuous cochains $C^n = C^n(G,M)$ as the group of continuous maps $G^n \rightarrow M$. We define a coboundary map $d_n : C^n \rightarrow C^{n+1}$ by the formula: $$ (d_n f)(g_1, \ldots, g_{n+1}) = g_1 f(g_2, \ldots, g_{n+1}) + \sum_{i=1}^n (-1)^i f(g_1, \ldots, g_ig_{i+1}, \ldots, g_{n+1}) + (-1)^{n+1} f(g_1, \ldots, g_n). $$

This yields a complex $C^{\bullet}(G,M)$. We define the $n$th cohomology group of $G$ with coefficients in $M$ by $H^n(G,M) := \text{ker } d_n / \text{im } d_{n-1}$.

Now, let $K$ be a finite extension of $\mathbb{Q}_p$ and $K^s$ a separable closure of $K$. Put $G_K:=Gal(K^s/K)$. Let $V$ be a finite-dimensional $\mathbb{Q}_p$-vector space with a continous action of $G_K$ given by $ρ:G_K \rightarrow GL(V)$.

My question is: Is there a difference between the cohomology groups $H^n(G_K,V)$ and $H^n(G_V,V)$, where $G_V=ρ(G_K)$?

I am guessing that if there is such a difference, it might have something to do with their respective topologies, (profinite topology on $G_K$ and $p$-adic topology on $GL(V)$) but I am not really sure.

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    $\begingroup$ Your definition of the relevant group cohomology is "wrong" insofar as one should make a definition that brings in the topologies under consideration. Using the correct definition, there is a huge difference; just try $n = 1$ and $\rho$ the trivial action on $V$ (so $G_V=1$). But in fact the topologies have nothing to do with the real origin of the distinction, since the same phenomenon occurs for cohomology of finite groups (acting on torsion abelian groups), where topology is irrelevant. $\endgroup$
    – user36938
    Jul 14, 2013 at 1:54
  • $\begingroup$ Thanks for pointing that out and for the idea. I already fixed the definition. $\endgroup$
    – Octobris
    Jul 14, 2013 at 3:23
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    $\begingroup$ Please treat homological algebra with more care and respect. Your "fix" does not make sense (away from the setting of discrete modules) since there's no reason why the category of topological modules for a topological group admits enough injectives (so it is not clear why derived functors in the way you're suggesting exist or are relevant to the correct definition -- one can make a correct definition in a couple of ways, such as via continuous cochains or inverse limits from discrete coefficients). $\endgroup$
    – user36938
    Jul 14, 2013 at 3:44
  • $\begingroup$ Please excuse my impertinence towards homological algebra. I hope the latest fix is acceptable enough. $\endgroup$
    – Octobris
    Jul 14, 2013 at 7:17
  • $\begingroup$ The fix looks fine. So now it is time for you to consider the case when the $G$-action on $V$ is trivial (so $G_V = 1$) and $n=1$, to figure out the answer to your own question (and to realize that the same phenomenon arises for cohomology of finite groups acting on torsion modules, so the topology and $p$-adic aspects are irrelevant to the key issues). $\endgroup$
    – user36938
    Jul 16, 2013 at 1:30

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